What is the solution for y^{''}+5y^'+4y=6e^{-x}+cos(2x) ?

The solution for y^{''}+5y^'+4y=6e^{-x}+cos(2x) is y=c_{1}e^{-x}+c_{2}e^{-4x}+2xe^{-x}+1/10 sin(2x)

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y^{''}+5y^'+4y=6e^{-x}+cos(2x)

y^{^{''}} + 5 y^{^{'}} + 4 y = 6 e^{- x} + cos (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x e^{- x} + \frac{1}{10} sin (2 x)

Solution steps

y^{''} (x) + 5 y^{'} (x) + 4 y = 6 e^{- x} + cos (2 x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x e^{- x} + \frac{1}{10} sin (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x e^{- x} + \frac{1}{10} sin (2 x)

y^{^{''}} + 4 y = t^{2} + 8 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} + 49 y^{^{'}} = tan (7 x)

y^{^{''}} + a y^{^{'}} = e^{3 t}

4 y^{^{''}} - 11 y^{^{'}} - 3 y = 5 e^{3 x}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x}

What is the solution for y^{''}+5y^'+4y=6e^{-x}+cos(2x) ?
The solution for y^{''}+5y^'+4y=6e^{-x}+cos(2x) is y=c_{1}e^{-x}+c_{2}e^{-4x}+2xe^{-x}+1/10 sin(2x)